Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x-3y &= -3 \\ -3x-8y &= -5\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-8y = 3x-5$ Divide both sides by $-8$ to isolate $y$ $y = {-\dfrac{3}{8}x + \dfrac{5}{8}}$ Substitute this expression for $y$ in the first equation. $3x-3({-\dfrac{3}{8}x + \dfrac{5}{8}}) = -3$ $3x + \dfrac{9}{8}x - \dfrac{15}{8} = -3$ Simplify by combining terms, then solve for $x$ $\dfrac{33}{8}x - \dfrac{15}{8} = -3$ $\dfrac{33}{8}x = -\dfrac{9}{8}$ $x = -\dfrac{3}{11}$ Substitute $-\dfrac{3}{11}$ for $x$ back into the top equation. $3( -\dfrac{3}{11})-3y = -3$ $-\dfrac{9}{11}-3y = -3$ $-3y = -\dfrac{24}{11}$ $y = \dfrac{8}{11}$ The solution is $\enspace x = -\dfrac{3}{11}, \enspace y = \dfrac{8}{11}$.